Born-Haber Cycle

Ionic solids tend to be very stable compounds. The enthalpies of formation of the ionic molecules cannot alone account for this stability. These compounds have an additional stability due to the lattice energy of the solid structure. However, lattice energy cannot be directly measured. The Born-Haber cycle allows us to understand and determine the lattice energies of ionic solids.

1. 1. Introduction
2. 2. Lattice Energy
3. 3. Born-Haber Cycle
4. 4. References
5. 5. Outside Links
6. 6. Problems
7. 7. Solutions
8. 8. Contributors

Introduction

This module will introduce the idea of lattice energy, as well as one process that allows us to calculate it: the Born-Haber Cycle. In order to use the Born-Haber Cycle, there are several concepts that we must understand first.

Lattice Energy

What is Lattice Energy?

Lattice Energy is a type of potential energy that may be defined in two ways. In one definition, the lattice energy is the energy required to break apart an ionic solid and convert its component atoms into gaseous ions. This definition causes the value for the lattice energy to always be positive, since this will always be an endothermic reaction. The other definition says that lattice energy is the reverse process, meaning it is the energy released when gaseous ions bind to form an ionic solid. As implied in the definition, this process will always be exothermic, and thus the value for lattice energy will be negative. Its values are usually expressed with the units kJ/mol.

Implications of Lattice Energy

Lattice Energy is used to explain the stability of ionic solids. Some might expect such an ordered structure to be less stable because the entropy of the system would be low. However, the crystalline structure allows each ion to interact with multiple oppositely charge ions, which causes a highly favorable change in the enthalpy of the system. A lot of energy is released as the oppositely charged ions interact. It is this that causes ionic solids to have such high melting and boiling points. Some require such high temperatures that they decompose before they can reach a melting and/or boiling point.

Born-Haber Cycle

Important Concepts

There are several important concept to understand before the Born-Haber Cycle can be applied to determine the lattice energy of an ionic solid; ionization energy, electron affinity, dissociation energy, sublimation energy, heat of formation, and Hess' Law.

Ionization Energy is the energy required to remove an electron from a neutral atom or an ion. This process always requires an input of energy, and thus will always have a positive value. In general, ionization energy increases across the periodic table from left to right, and decreases from top to bottom. There are some excepts, usually due to the stability of half-filled and completely filled orbitals.

Electron Affinity is the energy released when an electron is added to a neutral atom or an ion. Usually, energy released would have a negative value, but due to the definition of electron affinity, it is written as a positive value in most tables. Therefore, when used in calculating the lattice energy, we must remember to subtract the electron affinity, not add it. In general, electron affinity increases from left to right across the periodic table and decreases from top to bottom.

Dissociation energy is the energy required to break apart a compound. The dissociation of a compound is always an endothermic process, meaning it will always require an input of energy. Therefore, the change in energy is always positive. The magnitude of the dissociation energy depends on the electronegativity of the atoms involved.

Sublimation energy is the energy required to cause a change of phase from solid to gas, bypassing the liquid phase. This is an input of energy, and thus has a positive value. It may also be referred to as the energy of atomization. 

The heat of formation is the change in energy when forming a compound from its elements. This may be positive or negative, depending on the atoms involved and how they interact.

Hess' Law states that the overall change in energy of a process can be determined by breaking the process down into steps, then adding the changes in energy of each step. The Born-Haber Cycle is Hess' Law applied to an ionic solid.

Using the Born-Haber Cycle
The values used in the Born-Haber Cycle are all predetermined changes in enthalpy for the processes described in the section above. Hess' Law allows us to add or subtract these values, which allows us to determine the lattice energy.

Step 1:

Determine the energy of the metal and nonmetal in their elemental forms. (Elements in their natural state have an energy level of zero.) Subtract from this the heat of formation of the ionic solid that would be formed from combining these elements in the appropriate ration. This is the energy of the ionic solid, and will be used at the end of the process to determine the lattice energy.

Step 2:

The Born-Haber Cycle requires that the elements involved in the reaction are in their gaseous forms. Add the changes in enthalpy to turn one of the elements into its gaseous state, and then do the same for the other element.

Step 3:

Metals exist in nature as single atoms and thus no dissociation energy needs to be added for this element. However, many nonmetals will exist as polyatomic species. For example, Cl exists as Cl₂ in its elemental state. The energy required to change Cl₂ into 2Cl atoms must be added to the value obtained in Step 2.

Step 4:

Both the metal and nonmetal now need to be changed into their ionic forms, as they would exist in the ionic solid. To do this, the ionization energy of the metal will be added to the value from Step 3. Next, the electron affinity of the nonmetal will be subtracted from the previous value. It is subtracted because it is a release of energy associated with the addition of an electron. 

*This is a common error due to confusion caused by the definition of electron affinity, so be careful when doing this calculation.

Step 5:

Now the metal and nonmetal will be combined to form the ionic solid. This will cause a release of energy, which is called the lattice energy. The value for the lattice energy is the difference between the value from Step 1 and the value from Step 4.

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The diagram below is another representation of the Born-Haber Cycle.

Equation
The Born-Haber Cycle can be reduced to a single equation:

Heat of formation= Heat of atomization+ Dissociation energy+ (sum of Ionization energies)+ (sum of Electron affinities)+ Lattice energy

*Note: In this general equation, the electron affinity is added. However, when plugging in a value, determine whether energy is released (exothermic reaction) or absorbed (endothermic reaction) for each electron affinity. If energy is released, put a negative sign in front of the value; if energy is absorbed, the value should be positive.
Rearrangement to solve for lattice energy gives the equation:

Lattice energy= Heat of formation- Heat of atomization- Dissociation energy- (sum of Ionization energies)- (sum of Electron Affinities)

HESS’S LAW

Hess' Law is named after Russian Chemist and Doctor Germain Hess. Hess helped formulate the early principles of thermochemistry. His most famous paper, which was published in 1840, included his law on thermochemistry. The law went on to be called Hess' Law of Constant Heat summation (sometimes referred to as Hess' Law) and it states that regardless of the multiple stages or steps of a reaction, enthalpy change overall encompasses the sum of all changes.

1. 1. Introduction
2. 2. Definitions
3. 3. Why it works
4. 4. Steps involved in solving enthalpy of combustion problems: 
5. 5. Example
6. 6. Standard Enthalpy of Formation
7. 7. Standard Gibb's Energy of Formation
8. 8. Standard Entropy
9. 9. Worked Example
   1. 9.1. Steps
10. 10. Outside Links
11. 11. References 
12. 12. Contributors

Introduction

Hess' law is due to enthalpy being a state function, which allows us to calculate the overall change in enthalpy by simply summing up the changes for each step of the way, until product is formed. All steps have to proceed at the same temperature and the equations for the individual steps must balance out. Hess' law doesn't just apply to Enthalpy, it can also be used to calculate:

• Standard Gibb's Energy of Formation
• Standard Reaction Entropy 

Definitions

• The enthalpy of a given chemical reaction is constant, regardless of the reaction happening in one step or many steps.
• If a chemical equation can be written as the sum of several other chemical equations, the enthalpy change of the first chemical equation equals the sum of the enthalpy changes of the other chemical equations.

Why it works

The enthalpy of a reaction does not depend on the elementary steps, but on the final state of the products and initial state of the reactants.

• Enthalpy is an extensive property. An extensive property is a property that changes when the size of the sample changes. This means that the enthalpy of the reaction scales in a predictable manner. Enthalpy of reaction scales proportionally to the moles used in the reaction. For instance, in the following reaction, one can see that doubling the molar amounts simply doubles the enthalpy of the reaction.

H₂ (g) + 1/2O₂ (g) --> H₂O (g) ΔH° = -572 kJ

2H₂ (g) + O₂ (g) --> 2H₂O (g) ΔH° = -1144kJ

• The sign of the reaction enthalpy changes when a process is reversed.

H₂ (g) + 1/2O₂ (g) --> H₂O (g) ΔH° = -572 kJ

When switched: H₂O (g) --> H₂ (g) + 1/2O₂ (g) ΔH° = +572 kJ

• Since enthalpy is a state function, it is path independent. Therefor, it does not matter what reactions one uses to obtain the final reaction.

Steps involved in solving enthalpy of combustion problems: 

1. Balance the individual equations
2. If necessary look up standard enthalpies
3. Flip equations around if necessary to cancel out terms on opposite sides
4. Changing the equation around requires a sign change of the H of that individual step
5. Sum up the individual steps 

Example

Calculate the standard enthalpy of combustion of the transition of C(s, graphite) ==> C(s, diamond), given

C(s, graphite) + O₂ ==> CO₂ ΔH^o = -393.5 kJ/mol

CO₂ ==> C(s, diamond) + O₂ ΔH^o = + 395.41 kJ/mol

First we see that both equations are balanced. The enthalpies were given and there is no need to flip an equation around because it is possible to cancel out a couple terms as is. What is left is canceling out the O₂'s and the CO₂'s, writing the overall reaction and then summing the two enthalpies together.

C(s, graphite) + O₂ ==> CO₂

CO₂ ==> C(s, diamond) + O₂

Overall Equation becomes: C(s, graphite) ==> C(s, diamond)

• adding the enthalpies gives (-393.5 kJ/mol + 395.41 kJ/mol) = + 1.91 kJ/mol

Since the H^o is positive, this means that the reaction is endothermic.

Standard Enthalpy of Formation

The following equation is used to calculate the standard enthalpies of formation:

If H^o is positive ==> reaction is endothermic 

If H^o is negative ==> reaction is exothermic

• For concepts and information on Enthalpy. 

Standard Gibb's Energy of Formation

As mentioned above, the Standard Gibb's Energy of Formation is calculated using the principles of Hess's law. The following equation is used for those calculations:

To enlighten yourself about Gibb's energy click Gibb's Free Energy

Standard Entropy

The Standard entropy of a reaction is calculated using the following equation:

To learn the concepts and much more about Entropy.

Worked Example

Using Hess’s Law, the enthalpy of reaction of the major process of steam reforming can be determined.

CH₄(g) + H₂O(l) --> CO(g) + 3H₂(g) ΔH° = ??

Using the two postulates, given enough information, we are able to solve the enthalpy of reaction of an untabulated equation.

Using the reaction of Carbon Dioxide and Hydrogen gas and reaction of methane decomposition;

1) CO(g) + H₂(g) --> C(graphite) + H₂O(g) ΔH° = -131.3 kJ

2) C(graphite) + 2H₂(g) --> CH₄(g) ΔH° = -74.8 kJ

Steps

1. Flip Equation #2 to and change the sign of the enthalpy reaction.
   1. Ex. CH₄(g) --> C(graphite) + 2H₂(g) ΔH° = +74.8
2. Flip Equation #1 as well in order to get the unnecessary equation parts to cancel out.
   1. C(graphite) + H₂O --> CO(g) + H₂₍g) ΔH° = +131.3 kJ
3. With the two equations packed together, notice the equations will cancel out and come out with the final equation. C(graphite’s) cancel out between the two equations.

C(graphite) + H₂O --> CO(g) + H₂(g) ΔH° = +131.3 kJ

CH₄(g) --> C(graphite) + 2H₂(g) ΔH° = +74.8

Final Equation:

CH₄(g) + H₂O --> CO + 3H₂ ΔH° = +74.8 kJ + 131.3 kJ

4. Add the final enthalpies to the two fixed up equations and receive the enthalpy of the uknown equation. ?H° = +216.1

ΔH° = 216.1 kJ

CH₄(g) + H₂O -> CO + 3H₂ ΔH° = 216.1 kJ

Calorimetry

Calorimetry is the process of measuring the amount of heat released or absorbed during a chemical reaction. Calorimetry is a way of determining whether or not a reaction is exothermic, releases heat, or endothermic, absorbs heat. Calorimetry is a large aspect of thermodynamics, temperature measurements, because it gives the ability to collect data under specific conditions. Calorimetry is also a large part of everyday life, it even controls the metabolic rates in people, and controls our own bodies heat. Calorimetry can take place in any closed container which does not exchange heat with the surrounding environment. Any such container which is closed off from the environment is called a calorimeter.

1. 1. Introduction
2. 2. Outside Links
3. 3. References
4. 4. Contributors

Introduction

To calculate the amount of heat released or absorbed by a reaction with calorimetry, the heat of reaction formula is used. For example, if 150 grams of lead at 100°C were placed in a calorimeter with 50 grams of water at 28.8°C and the resulting temperature of the mixture was 22°C. We know that the specific heat of water is 4.184 J/g °C and the specific heat of lead is 0.128 J/g °C. Using calorimetry we can calculate the amount of heat gained and lost by each part of the system, so in this example we could calculate the amount of heat lost/gained by the water, the lead and the calorimeter. The amount of heat lost/gained by any part of the reaction is only related to the grams of the compound, its specific heat and the change in temperature it goes through, so we can relate each of the three parts together in the formula below.

The formula is Q = C x T where: Q = the amount of heat gained or lost C = the specific heat × mass and T = The final temperature of the mixture – the initial temperature of the substance

The heats of the different parts of the system can always be added up to equal zero because the amount of heat gained by one substance is equal to the amount lost by another. In the lead and water example the relationship would be as follows:

Q_lead+ Q_water +Q_calorimeter = 0

This can be helpful for calculating the heats of specific substances because any specific part of the equation can be subtracted to the other side of the equation and will equal the negative sum of the other two. If the heat of the water and the lead is known then the equation could be rearrange as such to find the heat of the calorimeter.

-(Q_lead) = Q_calorimeter.

With these two formulas we can now find the heats for all three parts of the reaction of lead and water using the information we started out with. For lead we know that:

Mass = 150 g, the initial temperature (T_i) = 100°C, and the specific heat (Csp) = 0.128 J/g °C

For water, mass = 50 g, T_i= 22°C, and C_sp = 4.184 J/g °C

And for the mixture of the two the final temperature (T_f) = 28.8°C

So with the two equations relating to heat we can calculate the amount of heat lost/gained for each separate part of the mixture

To calculate the amount of heat lost/gained by the lead we can use the firs equation, Q = C x T, because we have all the information needed to do. Q is the unknown, C = 0.128 J/g °C x 150g (specific heat x mass), and T = 28.8?C - 100°C (T_f – T_i) so the final calculation would go:

Qlead = 0.128 J/g °C x 150g x (28.8°C - 100°C) = -1.37 x 103 J

It is important to see that the heat for lead is negative because the lead declined in temperature and lost heat when it was mixed with water. The water temperature did the opposite when the lead was added to it and rose which tells us that the water gained heat when it was mixed with the lead and because we have the same information to start with we can use the same information to calculate exactly how much heat was gained, so

QH20 = 4.184 J/g °C x 50g x (28.8°C - 22°C) = 1.42 x 103 J

Now that we have the amount of heat which the lead lost and the amount that the water gained, we can find how much heat the calorimeter lost/gained by using the relationship -(Qlead + Qwater) = Qcalorimeter., so

Q_cal = -(Q_lead + Q_H20) = -(1.42 x 103 +-1.37 x103) = -50.0 J

With those calculations we now know that when the water and lead were mixed inside the calorimeter, the lead lost -1.37 x 103 J of heat, the water gained 1.42 x 103 J of heat, and the calorimeter lost -50.0 J of heat.

Calorimetry is a technique to measure the amount of heat energy created or absorbed in a chemical process. An attempt is made to conserve all the evolved heat using a calorimeter, a device containing a suitable liquid that makes good thermal contact with the process under investigation. 

The following headers are meant as a guide to the sections that you need in your module. Here, before the table of contents, give a two-four sentence overview of the Module. Be straight to the point. No examples and no figures in this section.  Keep the following text for automatic table of contents based on the headings below (seen in edit mode only).

1. 1. Introduction
2. 2. Bomb and Coffee-Cup Calorimeters  
3. 3. References
4. 4. Outside Links
5. 5. Problems
6. 6.  

Introduction

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A calorimeter is a device used to measure the quantity of heat flow in a chemical reaction. Two of the most common types of calorimeters are the coffee cup calorimeter and the bomb calorimeter.

Bomb and Coffee-Cup Calorimeters  

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Coffee Cup Calorimeter
Coffee cup calorimeters are often utilized to demonstrate the concepts behind calorimetry. A known quantity of water is added into a polystyrene cup and a thermometer is inserted into the system so that the bulb is immersed in water. When a reaction occurs in this coffee cup calorimeter, the heat of the reaction will cause the water temperature to change.
This change  in water temperature is indicative of the amount of heat absorbed
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References

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Important Terms in Calorimeter

 

•Specific heat capacity, c

–Specific heat capacity, c of a substance is the amount of heat required to raise the temperature of one gram of the substance by one degree Celsius (Jg -1°C-1).

•Heat capacity, C

–Heat capacity,C is the amount of heat required to raise the temperature of a given quantity of the substance by one degree Celsius (J°C-1)

Enthalpy of Sublimation, Hsubl

The heat (or enthalpy) of sublimation is the amount of energy that must be added to a solid at constant pressure in order to turn it directly into a gas (without passing through the liquid phase).  The heat of sublimation is generally expressed as ΔH_sub in units of Joules or kiloJoules per mole or kilogram of substance.

1. 1. Things to recall...
2. 2. Introduction to Sublimation and Heat of Sublimation
3. 3. The Equations
   1. 3.1. ΔHsub = ΔEtot
      1. 3.1.1. ΔEthermal(state of matter)
      2. 3.1.2. ΔEbond(going from state 1 to state 2)
4. 4. Graphical Representations of the Heat of Sublimation
5. 5. ΔHsub > ΔHvap
6. 6. Where does the added energy go?
7. 7. Practical Applications of the Heat of Sublimation
8. 8. Practice Problems
9. 9. Practice Problem Solutions
10. 10. Footnotes
11. 11. External References
12. 12. Contributors

Things to recall...

1. A "Δ" in front of any variable indicates that the following equation is a state function in which the value of an indicator measured in the initial state of a reaction is subtracted from the value of the same indicator measured in the final state of the reaction.  The resulting value is the change in the indicator at the end of the said reaction (Δindicator). 
2. Enthalpy is the amount of energy that is required to induce a phase change (change in the state of matter).
3. Energy is measured in Joules or kiloJoules and is transferred through either through heat (q) or work (w).
4. The phases involved in sublimation are the solid phase and the gas phase.

Introduction to Sublimation and Heat of Sublimation

Sublimation is the process of changing a solid into a gas without allowing the solid to pass through the liquid phase.  In order to sublime a substance, a certain amount of energy must be transferred to the substance via heat (q) or work (w).  The energy needed to sublime a substance is particular to the substance's identity and temperature and must be enough to do all of the following:

1. Excite the substance_(s) so that it reaches its maximum heat (energy) capacity (q) in the solid state.
2. Sever all the atomic bonds in the substance_(s).
3. Excite the unbonded atoms of the substance so that it reaches its minimum heat capacity in the gaseous state.

(See attached powerpoint show in the Files section at the bottom of the page.  Select the option to play slideshow.)

The Equations

ΔH_{sub =} ΔE_tot

The energies involved in sublimation steps 1 through 3 (see Sublimation ) compose the total amount of energy that is involved in sublimation.  This can be expressed by the equation,

in which

Although a solid does not actually pass through the liquid phase during the process of sublimation, the fact that enthalpy is a state function allows us to add the various energies associated with the solid, liquid, and gas phases together.  Recall that for state functions, only the initial and final states of the substance are important.  Say for example that state A is the initial state and state B is the final state.  How a substance goes from state A to state B does not matter so much as what state A and what state B are.  Concerning the state function of enthalpy, the energies associated with enthalpies (whose associated states of matter are contiguous to one another) are additive.  Though in sublimation a solid does not pass through the liquid phase on its way to the gas phase, it takes the same amount of energy that it would to first melt (fuse) and then vaporize. 

ΔE_{thermal(state of matter)}

A change in thermal energy is indicated by a change in temperature (in Kelvin) of a substance at any particular state of matter.  Change in thermal energy is expressed by the equation

in which

For more information on heat capacity and specific heat capacity, see heat capacity.  Specific heat capacities of common substances can easily be found online or in a reference or text book.

ΔE_{bond(going from state 1 to state 2)}

Bond energy is the amount of energy that a group of atoms must absorb so that it can undergo a phase change (going from a state of lower energy to a state of higher energy).  It is measured

in which is the enthalpy associated with a specific substance at a specific phase change.  Common types of enthalpies include the heat of fusion (melting) and the heat of vaporization.  Recall that fusion is the phase change that occurs between the solid state and the liquid state, and vaporization is the phase change that occurs between the liquid state and the gas state.  Note that if the substance has more than one type of bond (or intramolecular force), then the substance must absorb enough energy to break all the different types of bonds before the substance can sublime.¹

Graphical Representations of the Heat of Sublimation

Note that although the graph below indicates the inclusion of the liquid phase, the graph is merely a representation of how much energy is needed to sublime a solid substance.  Recall that sublimation does not include the liquid phase and that the fact that enthalpy is a state function allows us to add the enthalpies of fusion and vaporization together to find the enthalpy of sublimation.

        

ΔH_{sub >} ΔH_vap

Though both enthalpies involve the changing a substance into its gaseous state, the change in energy associated with sublimation is generally greater than that of vaporization.  This is because of the initial state of the substances and the amount of initial energy that each substance has.  Particles in a solid have less energy than those of a liquid, meaning it is takes more energy to excite a solid to its gaseous phase that it does to excite a liquid to its gaseous phase. 

Another way to look this phenomena is to take a look at the different energies involved with the heat of sublimation:  ΔE_{thermal (s)}, ΔE_bond(s-l), ΔE_thermal(l), and ΔE_bond(l-g).  Already we know that ΔE_bond=ΔH_{(phase change)}*Δm_{(changed substance)} and ΔE_bond(l-g)=ΔH_(l-g)*Δm_{(gas created)}.  ΔH_(l-g) is essentially ΔH_{(vaporization)}, meaning that ΔH_vap is actually a component of ΔH_sub.

 

Where does the added energy go?

Energy can be observed in many different ways.  As shown above, ΔE_tot can be expressed as ΔE_{thermal +} ΔE_bond.  Another way in which  ΔE_tot can be expressed is change in potential energy, ΔPE, plus change in kinetic energy, ΔKE.  Potential energy is the energy associated with random movement, whereas kinetic energy is the energy associated with velocity (movement with direction).  ΔE_tot = ΔE_{thermal +} ΔE_bond and ΔE_tot = ΔPE + ΔKE are related by the equations

ΔPE = (0.5)ΔE_{thermal +} ΔE_bond ΔKE = (0.5)ΔE_thermal

for substances in the solid and liquid states.  Note that ΔE_thermal is divided equally between ΔPE and ΔKE for substances in the solid and liquid states.  This is because the intermolecular and intramolecular forces that exist between the atoms of the substance (i.e. atomic bond, van der Waals forces, etc) have not yet been dissociated and prevent the atomic particles from moving freely about the atmosphere (with velocity).  Potential energy is just a way to have energy, and it generally describes the random movement that occurs when atoms are forced to be close to one another.  Likewise, kinetic energy is just another way to have energy, which describes an atom's vigorous struggle to move and to break away from the group of atoms.  The thermal energy that is added to the substance is thus divided equally between the potential and the kinetic energies  because all aspects of the atoms' movement must be excited equally

However, once the intermolecular and intramolecular forces which restrict the atoms' movement are dissociated (when enough energy has been added), potential energy no longer exists (for monatomic gases) because the atoms of the substance are no longer forced to vibrate and be in contact with other atoms.  When a group of atoms is in the gaseous state, it's atoms can devote all their energies into moving away from one another (kinetic energy).

Enthalpy of Solution, Hsoln

The enthalpy of solution, enthalpy of dissolution, or heat of solution is the enthalpy change associated with the dissolution of a substance in a solvent at constant pressure resulting in infinite dilution.
The enthalpy of solution is one of the three dimensions of solubility analysis. It is most often expressed in kJ/mol at constant temperature. Just as the energy of forming a chemical bond is the difference between electron affinity and ionization energy, the heat of solution of a substance is defined as the sum of the energy absorbed, or endothermic energy, expressed in positive values and unit kJ/mol, and energy released, or exothermic energy (negative value).
Because heating decreases the solubility of a gas, dissolution of gases is exothermic. Consequently, as a gas continues to dissolve in a liquid solvent, temperature will decrease, while the solution continues to release energy. This is an effect of the increase in heat or of the energy required to attract solute and solvent molecules—in other words, this energy outweighs the energy required to separate solvent molecules. When the gas is completely dissolved (this is purely theoretical as no substance can infinitely dissolve), the heat of solution will be at its maximum.
Dissolution can be viewed as occurring in three steps:

1. Breaking solute-solute attractions (endothermic), see for instance lattice energy in salts.
2. Breaking solvent-solvent attractions (endothermic), for instance that of hydrogen bonding
3. Forming solvent-solute attractions (exothermic), in solvation.

The value of the overall enthalpy change is the sum of the individual enthalpy changes of each of these steps. For example, dissolving ammonium nitrate in water decreases the temperature of the solution. Solvation does not compensate energy spent in breaking down the crystal lattice, while adding potassium hydroxide will increase it.
Solutions with negative enthalpy changes of solution form stronger bonds and have lower vapor pressure.
The enthalpy of solution of an ideal solution is zero since the attractive and repulsive properties of ideal fluids are equal, irrespective of the compounds, thus mixing them does not change the interactions.

Enthalpy change of solution for some selected compounds
hydrochloric acid	-74.84
ammonium nitrate	+25.69
ammonia	-30.50
potassium hydroxide	-57.61
caesium hydroxide	-71.55
sodium chloride	+3.88
potassium chlorate	+41.38
acetic acid	-1.51
sodium hydroxide	-44.51
Change in enthalpy ΔHo in kJ/mol in water at 25°C[1]

Enthalpy of Hydration, Hhyd

•The heat change when 1 mole of gaseous ions is hydrated in water.

•e.g:

  Na+(g) ¾® Na+(aq)    DHhyd = -406 kJ mol-1

  Cl-(g) ¾® Cl-(aq)        DHhyd = -363 kJ mol-1

Hydration

The formation of a solution involves the interaction of solute with solvent molecules. Many different liquids can be used as solvents for liquid solutions, and water is the most commonly used solvent. When water is used as the solvent, the dissolving process is called hydration.

The interaction between water molecules and sodium ion is illustrated as one of the diagram below. This is a typical ion-dipole interaction. At the molecular level, the ions interact with water molecules from all directions in a 3-dimensional space. This diagram depicts the concept of interaction only.

The above diagram also display hydrogen-bonding, dipole-dipole, ion-induced dipole, and dipole-induced dipole interactions. In the absence of these interactions, solvation takes place due to dispersion. Definitions of these terms are obvious from the diagrams. The meaning of the words used in the term also hints the nature of the interactions.

What is the energy of hydration?

Enthalpy of hydration, H_hyd, of an ion is the amount of energy released when a mole of the ion dissolves in a large amount of water forming an infinite dilute solution in the process, M^z+(g) + mH₂O ® M^z+(aq) where M^z+(aq) represents ions surrounded by water molecules and dispersed in the solution. The approximate hydration energies of some typical ions are listed here. The table illustrates the point that as the atomic numbers increases, so do the ionic size, leading to a decrease in absolute values of enthalpy of hydration.

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Enthalpy of Hydration (H_hyd kJ/mol) of Some Typical Ions

Ion	Hhyd	Ion	Hhyd	Ion	Hhyd
H+	-1130	Al3+	-4665	Fe3+	-4430
--
Li+	-520	Be2+	-2494	F-	-505
Na+	-406	Mg2+	-1921	Cl-	-363
K+	-322	Ca2+	-1577	Br-	-336
Rb+	-297	Sr2+	-1443	I-	-295
Cs+	-276	Ba2+	-1305	ClO4-	-238
--
Cr2+	-1904	Mn2+	-1841	Fe2+	-1946
Co2+	-1996	Ni2+	-2105	Cu2+	-2100
Zn2+	-2046	Cd2+	-1807	Hg2+	-1824

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From the above table, an estimate can be made for the hydration energy of sodium chloride. This amount is very close to the energy of crystallization, E_cryst. The hydration energy of an ionic compound consists of two inseparable parts. The first part is the energy released when the solvent forms a coordination compound with the ions. This energy released is called the energy of ligation, H_lig. The processes related to these energies are shown below:
M^z+ + nL = ML_n^z+,             H_lig
ML_n^z+ + solvent = ML_n^z+ (solution),     H_disp The second part is to disperse the ions or hydrated ions into the solvent medium, which has a dielectric constant different from vacume. This amount of energy is called energy of dispersion, H_disp. Therefore, H_hyd = H_disp + H_lig. This idea is brought up just to point out that the formation of aqua complex ions is part of the hydration process, even though the two energies are not separable. When stronger coordination is made between the ions and other ligands, they replace the coordinated water molecules if they are present. In the presence of NH₃ molecules, they replace the water of Cu(H₂O)₆²⁺: Cu(H₂O)₆²⁺ + 6NH₃ ® Cu(NH₃)₆²⁺ + 6H₂O

How is hydration energy related to enthalpy of crystallization?

In the discussion of lattice energy, we consider the ions separated into a gas form whereas in the disolution process, the ions are also separated, but this time into ions dispersed in a medium with solvent molecules between ions. The medium or solvent has a dielectric constant. The molar enthalpy of solvation, H_solv, is the energy released when one mole solid is dissolved in a solvent. This quantity, the enthalpy of crystallization, and energy of hydration forms a cycle. Taking the salt NaCl as an example, the following relationship is obvious, H_hyd = H_{crystallization} + H_solv from the following diagram.

         -------Na+(g)+Cl-(g)--------
         |                       |
         |                       |
         |                       |Hcryst
         |Hhyd                    |
         |                       ¯
         |                 ----NaCl(s)---
         |                       |
         |                       |Hsolv
         ¯                       ¯
         -------Na+(aq)+Cl-(aq)--------

The term enthalpy of crystallization is used in this diagram instead of lattice energy so that all the arrows point downward. Note that enthalpy of crystallization H_cryst, and energy of crystallization, E_cryst referr to the same quantity, and they are used interchangably. The energies of solvation for some salts can be positive values, in these cases the temperatures of the solution decrease as the substances dissolve. The solvation is an endothermic reaction. The energy levels of solids and solutions reverse in order of hight. The cycle is shown below.

         -------Na+(g)+Cl-(g)-------
         |                       |
         |                       |
         |                       |Hcryst
         |Hhyd                    |
         |                       |
         ¯                       |
         ---Na+(aq)+Cl-(aq )----    |
                            ­    |
                      Hsolv  |    |
                            |    ¯
                   ---------NaCl(s)--------

In these cases, the enthalpies of hydration are less negative than the enthalpies of crystallization.

What is enthalpy of solvation?

The molar enthalpy of solvation, H_solv, is the energy released when one mole solid is dissolved in a solvent. Note that H_solv is also called molar heat of solution or molar energy of solution in some literature. Sometimes the enthalpy of hydration is also (mis)understood as H_solv. When apply these values, make sure you understand the process involved. The following enthalpies of solvation are given in Chemistry by Radel and Navidi, West Publishing Co.

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Enthalpy of Solvation (H_solv kJ/mol) of Some Common Electrolytes

Substanc	Hsolv	Substance	Hsolv
AlCl3(s)	-373.63	H2SO4(l)	-95.28
LiNO3(s)	-2.51	LiCl(s)	-37.03
NaNO3(s)	20.50	NaCl(s)	3.88
KNO3(s)	34.89	KCl(s)	-17.22
NaOH(s)	-44.51	NH4Cl(s)	14.77

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These values indicates that when aluminum chloride and sulfuric acid are dissolved in water, much heat is released. Due to the very small value of enthalpies of solvation, the temperature changes are hardly noticed when LiNO₃ and NaCl are dissolving.